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Requirements for inductor to operate in continuous conduction mode? 
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#1
Mar2013, 07:43 PM

P: 102

I have a few quesitons about CCM converters
I have posted a basic schematic of the circuit in question as an attachment. The circuit is a series LC circuit in which the inductor and capacitor charge during the on time. During the off time the inductor energy dumps into the cap, charging it to 2x the supply voltage. My questions about the circuit:  What would be required to get the circuit to operate in CCM? Since the charging/discharging impedances are equal could a 50% duty cycle be used? Would the circuit only operate in CCM when the pulsing frequency is equal to the resonant time period between the inductor and capacitor? Any help would be greatly appreciated! Thank you. 


#2
Mar2113, 08:38 AM

Sci Advisor
PF Gold
P: 3,503

no help yet?
see if there's anything helpful here. http://www.coilcraft.com/prod_pwr.cfm one of the major manufacturers had tutorials on switching regulator design , I think it was National before TI bought them. Try a search on phrase "Simple Switchers" 


#3
Mar2113, 05:25 PM

P: 102

That site has lots of good info but not exactly what I am looking for.
I have the simulation circuit (attached to my first post). All I want to understand is what the requirements are to get the inductor to operate in continuous mode when the input is a 50% duty cycle pulse. 


#4
Mar2113, 06:12 PM

Mentor
P: 40,649

Requirements for inductor to operate in continuous conduction mode?



#5
Mar2113, 06:15 PM

Mentor
P: 40,649

The way you are simulating the boost circuit is a bit nontraditional, BTW. This is a more traditional way to make a boost circuit:
http://ars.elscdn.com/content/image...001488gr3.jpg 


#6
Mar2113, 09:48 PM

P: 102

So, the average output current has to be over half the ripple current.
Can you explain why this is? 


#7
Mar2113, 10:04 PM

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P: 40,649

For continuous output current mode, you do not want to drop to zero output current at the bottom of the ripple current waveform. So if the average output current is at least half of that ripple current, the graph of the output current waveform never reaches the bottom axis (I=0). The main problem with allowing nonCCM is that it complicates the feedback equations, and makes it harder to design a stable DCDC converter circuit. Not impossible, of course, but it makes it harder, and constrains the performance of the circuit (either you add more circuitry to make it more expensive, or you settle for reduced regulator performance). 


#8
Mar2113, 10:24 PM

P: 102

Ok, that makes sense.
Thanks! I appreciate the help. 


#9
Mar2213, 01:05 PM

P: 384

Looking at the circuit that hobbs125 gave in his original post, isn't the determination of whether the inductor current is continuous or discontinuous dependent on the resonant frequency of L1C1 and the frequency of the square wave?
(This is a guess on my part.) This would be the ideal case for a spice simulation. Is there anyone here that can do a spice simulation? 


#10
Mar2213, 01:24 PM

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#11
Mar2313, 11:44 AM

P: 384

Quote from original post "The circuit is a series LC circuit in which the inductor and capacitor charge during the on time. During the off time the inductor energy dumps into the cap, charging it to 2x the supply voltage."
The preceding is only partially correct, the capacitor is only charged to 2x the supply voltage if the capacitor is charged to zero volts at start of pulse. In "normal" operation the capacitor's average voltage would be slightly less than the square wave pulse peak voltage. Using a 1.13 mHy inductor, a 1N4148 diode, 5 each 2.2 uf capacitors 1.12 uf measured and a 1000 ohm resistor + a function generator and oscilloscope. With a 20 volt peak to peak square wave input, the start of the CCM was measured at between 500 kHz and 1.0 MHz. (Depending on how the oscilloscope wave form was interpreted.) It would be interesting if someone would go through the math and say if the measure frequency for CCM is the same as the calculated CCM. 


#12
Mar2313, 02:05 PM

P: 102

Carl Pugh,
This is exactly what I am wondering....My question, is why? What is different at Fres that causes the circuit to operate in CCM? Then again, I think berkeman has a good point. I tested the circuit in multisim and it seems to operate in CCM at all frequencies..But the resonant thing really has me wondering still, not sure why? It just seems like we would see something different occuring at that freq, even though it's not an AC circuit. 


#13
Mar2313, 11:41 PM

P: 102

Carl



#14
Mar2413, 12:43 PM

P: 384

What is the definition for a square wave?
Whether the circuit is CCM or not depends at least partially on the definition of a square wave. 


#15
Mar2413, 06:50 PM

P: 102

Well, those are all square waves as far as I know. However, the circuit I posted contains a rectifier and is meant for unipolar pulses.
I do know that if you breakdown the square wave you find out it's made of multiple harmonic AC waves.....IS that what your getting at? 


#16
Mar2413, 09:30 PM

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P: 40,649




#17
Mar2413, 09:45 PM

P: 102

This circuit is more of an inductive charging circuit than a boost converter, meant to charge the cap to 2x Vs. Problem is, the load drains the cap too fast and causes the voltage to drop. That's why I want to know about how to get the cirucit to operate in ccm, if there's a certain frequency required and how to calculate that frequency? I still don't really understand it. 


#18
Mar2513, 11:52 AM

P: 384

Using the circuit and component values originally posted, assuming that all components are ideal and that the square wave goes between 0 and +10 volt.
At high frequencies, the inductor acts as an open circuit to AC and the capacitor only charges to the average voltage of the square wave, or 5 VDC. The current through the inductor goes continuous when the frequency of the square wave increases to a high enough frequency that the capacitor voltage drops to 5 VDC THE FOLLOWING CALCULATES THE LOWEST FREQUENCY AT WHICH CCM OCCURS For positive portion of square wave, di/dt = (voltage of square waveC1 voltage)/Inductance di/dt = (105)/0.0017 = 2940 amp/second For 0 voltage portion of square wave, di/dt = (voltage of square waveC1 voltage)/Inductance di/dt = (05)/.0017 = 2940 amp/second The current through the resistor is 5 volt/1000 ohm = 0.005 amp For the average current through the inductor to be 0.005 amp, the peak current has to be twice this or 0.01 amp. 0.01 amp/2940 amp/second=3.40 microsecond that the square wave is at 10 volt. Since the di/dt is the same for when the square wave is at 0 volt, the period of the square wave is 2 X 3.4 microsecond or 6.8 microsecond. f = 1/period = 1/6.8 microsecond = 147 kilohertz At frequencies below 147 kilohertz, the current through L1 is discontinuous. At frequencies above 147 kilohertz, the current through L1 is continuous. C1 never charges to a higher voltage than the peak voltage of the square wave. When L1 is CCM, the output voltage is one half the square wave peak voltage. (Tested this circuit using component values I gave previously and the two preceding statements are correct) 


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